Co Ordination Compounds Question 51
Question: $ M{n^{2+}} $ forms a complex with $ B{r^{-}} $ ion. The magnetic moment of the complex is 5.92 B.M. What would be the probable formula and geometry of the complex?
Options:
A) $ {{[MnBr_6]}^{4-}} $ , octahedral
B) $ {{[MnBr_4]}^{2-}} $ , square planar
C) $ {{[MnBr_4]}^{2-}} $ , tetrahedral
D) $ {{[MnBr_5]}^{3-}} $ , trigonal bipyramidal
Show Answer
Answer:
Correct Answer: C
Solution:
Magnetic moment, $ \sqrt{n(n+2)}=5.92,B.M.$
$ \Rightarrow n $ (unpaired electrons) = 5 $ sp^{3} $ hybridization (tetrahedral)
BETA
